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  • Obtain the binding energy of the nuclei 56 26 Fe and 209 83 Bi in units of MeV from the following data: m (26 56 Fe) = 55.934939 u

Q. 13.4 (i) Obtain the binding energy of the nuclei _{26}^{56}\textrm{Fe} and _{83}^{209}\textrm{Bi} in units of MeV from the following data:

   (i)m (_{26}^{56}\textrm{Fe})=55.934939\; \; u
                

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mH = 1.007825 u

mn = 1.008665 u

Atomic mass of _{26}^{56}\textrm{Fe} is m=55.934939 u

Mass defect 

\Delta m=(56-26)\timesm_H+26\times m_p - m

\Delta m=30\times1.008665+26\times1.007825 - 55.934939

\Deltam=0.528461

Now 1u is equivalent to 931.5 MeV

Eb=0.528461\times931.5

Eb=492.2614215 MeV

Therefore binding energy of a _{26}^{56}\textrm{Fe} nucleus is 492.2614215 MeV.

Average binding energy

=\frac{492.26}{56}MeV=8.79 MeV

Posted by

Sayak

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